What my undergraduate thesis (Poonen, Maximally complete fields, Enseign. Math. (2) 39 (1993), no. 1-2, 87-106) did was to generalize the $p$-adic construction to arbitrary value groups and residue fields, to prove that $\Omega_p$ and these generalizations are maximally complete (hence spherically complete, hence the same as Kaplansky's fields), and to prove various other statements - for instance, that the transcendence degree of $\Omega_p$ over $\mathbb{C}_p$ is $2^{\aleph_0}$.

Much of the general theory of spherically complete fields was developed in the 1930s and 1940s by Schmidt, Krull, Ostrowski, and especially Kaplansky in his thesis doi.org/10.1215/S0012-7094-42-00922-0 . The explicit description of the field of transfinite series is due to Hahn in 1908 in the equicharacteristic case and to Lampert doi.org/10.1016/0022-314X(86)90073-9 in 1986 in the $p$-adic case above.

@Rdrr: Here is another way to say it. Let $\mathfrak{p}$ be the prime corresponding to $v$. The implication $v(\beta) \ge 2n \implies \beta \in (p)$ is equivalent to $\mathfrak{p}^{2n} \subset (p)$, which is the same as saying that $(p)$ divides $\mathfrak{p}^{2n}$. In this case, the factorization of the $\mathcal{O}$-ideal $(p)$ cannot involve any primes other than $\mathfrak{p}$.

When I wrote "that rule is not exact", I meant specifically that the functor sending $G$ to $\operatorname{Map}(G^*,\mathbf{G}_m)$ is not exact, so certainly the functor sending $G$ to the Mazur-Roberts sequence $0 \to G \to A \to B \to 0$ is not exact.

I think that that rule is not exact. It sends $1 \to \mu_2 \to \mu_6 \to \mu_3 \to 1$ to $1 \to \mathbf{G}_m^2 \to \mathbf{G}_m^6 \to \mathbf{G}_m^3 \to 1$, and $2 + 3 \ne 6$. Anyway, the pushout argument is not hard: Choose $G' \hookrightarrow A'$ and $G \hookrightarrow \mathcal{A}$; then $0 \to G' \to G \to G/G' \to 0$ embeds in $0 \to A' \to (A' \times \mathcal{A})/G' \to \mathcal{A}/G' \to 0$ (with $G'$ in the quotient in the middle being the antidiagonal copy), and then one can take cokernels to get $0 \to B' \to B \to B'' \to 0$.

In contrast to what is claimed above, deciding whether a two-variable polynomial equation has an integer solution is still an open problem. Runge's method (used in the Hilliker-Straus paper mentioned above) and Baker's method are effective only for special equations.

To prove that every finite abelian group is a Galois group over $\mathbf{Q}$, one does not even need Dirchlet's theorem on primes in arithmetic progression: the argument in the answer to mathoverflow.net/questions/15220/… suffices.

An algorithm for two-variable equations of degree <=3 already exists! WLOG the polynomial is irreducible over Z, hence irreducible over Q. If it not absolutely irreducible, then the absolute factors are conjugate, so any rational solution to one of them satisfies them all, so one can solve the system to find the finitely many rational points and check whether any of them are integer points.

The main point of Theorem 1.2 in that Cadoret-Tamagawa paper was to bound $\dim J$ in terms of $\dim A$. The existence of $J$ is older, going back at least to Corollary 2.5 in this article of Gabber

@Yuan Yang: You could use Hartshorne III.9.10 instead, at least if you are willing to assume that $X$ is projective over $\mathbb{Z}$ (which is true: see mathoverflow.net/questions/207443 for an even more general statement).

@Yuan Yang: Smooth implies flat. The relative Jacobian ($\operatorname{Pic}^0$, not Pic) is an abelian scheme: WLOG the relative curve is connected; then the fibers are geometrically connected since there is no nontrivial finite étale cover of Spec $\mathbb{Z}$, so Prop. 4 on p. 260 of Bosch-Lütkebohmert-Raynaud applies. In your last sentence, in the higher-dimensional case, I will assume that by "trivial Picard scheme" you meant "trivial $\operatorname{Pic}^0$"; the argument seems incomplete, however, since it's not obvious a priori that $\operatorname{Pic}^0$ is an abelian scheme.

Also, ${\mathcal N}_{D/X}$ should be ${\mathcal N}_{D/Y}$, I suppose.

I believe that the relative tangent bundle of $f \colon X \to Y$ is the negative of the normal bundle. Also, $O_Y(D)$ above should be $f_* {\mathcal O}_D$, that is, ${\mathcal O}_D$ viewed as a coherent sheaf on $Y$, as appears in the exact sequence.

Your version is equivalent, but it seems to require a short argument combining two copies of $\mathcal{F}$ to show this. The original version has the advantage that it allows $\mathcal{F}$ to be the power set of a nonempty set $A$.